Let x be the concentration of H3O+ (mol/L) formed when HCO3- reacts. Since H2O is present in excess, we can assume that the initial concentration of HCO3- is equal to the concentration formed:

[HCO3-] = x (mol/L)

[H3O+] = x (mol/L)

The equation for the equilibrium constant (Ka) is given as:

Ka = [H3O+][CO32-] / [HCO3-]

Given Ka = 4.7 x 10^-11 and [HCO3-] = 8.127 x 10^17 molecules in 1.0 liter (which is equivalent to 8.127 x 10^-8 mol/L):

4.7 x 10^-11 = (x)(x) / (8.127 x 10^-8)

Solving for x:

x^2 = 4.7 x 10^-11 * 8.127 x 10^-8

x^2 = 3.82969 x 10^-18

x ≈ 1.956 x 10^-9 mol/L

Now, to find the pH, we can use the equation:

pH = -log[H3O+]

pH ≈ -log(1.956 x 10^-9)

pH ≈ 8.71

So, the pH of the ocean based on this sample is approximately 8.71.

Ionic equation for the reaction of Sodium Nitrate (NaNO3) with Potassium Carbonate (K2CO3) in water:

NaNO3(aq) + K2CO3(aq) → 2KNO3(aq) + Na2CO3(aq)

Ionic equation for the addition of Sulfurous Acid (H2SO3) to water:

H2SO3(aq) → H+(aq) + HSO3-(aq)